# Help with a geometry problem

The problem says: A triangle has its lengths in an arithmetic progression, with difference d. The area of the triangle is t. Find the dimensions.

the solution says: the notation can be even better if we make it more symmetrical, by making the side lengths $b − d, b, and\ b + d$ .

by Heron’s formula we know that $t^2 = s(s − b + d)(s − b)(s − b − d)$ , where $s = ((b − d) + b + (b + d))/2$ is the semi-perimeter;

and after simplification $$t^2 = \frac{3b}{2}(\frac{3b}{2} - b + d )(\frac{3b}{2} - b ) (\frac{3b}{2} - b - d )$$ $$\implies t^2 = \frac{3b^2(b-2d)(b+2d)}{16} = \frac{3b^2(b^2-4d^2)}{16}$$

then we get $$3b^4 − 12d^2b^2 − 16t^2 = 0$$

and using the quadratic formula : $$b^2 = \frac{12d^2 \pm \sqrt{144d^4 + 169t^2} }{6} = 2d^2 \pm \sqrt{4d^4 + \frac{16}{3} t^2}$$

and because b has to be positive , we get

$$b = \sqrt{2d^2 + \sqrt{4d^4 + \frac{16}{3}t^2}}$$

Which is the part that i have a problem with , my question is : why should we select only the positive sign solution of the quadratic formula ? is that because $\sqrt{ 4d^4 + \frac{16}{3}t^2} > 2d^2$ which means that the negative sign solution leads to the square root of a negative number which is not valid? why is the positive sign solution is the right solution ?

In other words :

If $\sqrt{ 4d^4 + \frac{16}{3}t^2} > 2d^2$ , how is that ? how can we prove it ?

thank you

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We have $4d^4=(2d^2)^2$, and therefore $4d^4+\frac{3}{16}t^2>(2d^2)^2$ and therefore $$\sqrt{4d^4+\frac{3}{16}t^2}>2d^2.$$ It follows that $2d^2-\sqrt{4d^4+\frac{3}{16}t^2}<0$, so $2d^2-\sqrt{4d^4+\frac{3}{16}t^2}$ does not have a (real) square root.
Yes, you know that $b^2$ and $b$ are both positive real numbers, so you select the positive sign twice while taking square root.